Projectile/ Trajectory equation

[articulate]

Hi guys,

I have some questions about this equation that my class were given before carrying out our practical for projectile motion (what we did was release a marble from the top of an inclined ramp and then when it hits the board which has a piece of carbon paper on it, a mark is left on a piece of paper under the carbon paper..)

Here's the equation:

y/x=[g(1+k^2)x]/2u^2 + k

k is a constant
g=gravitational acceleration
y=vertical distance (between bottom of ramp and top of board)
x=horizontal distance (between plumline and mark on paper)

Questions:

1. What is k really? (I read from some other sites that k is air resistance. Is that true?) From what equation was it derived from?
2. How do I get this equation?


What I've come up with so far..:

y=(u sin Ɵ)t + 1/2 gt^2

x= u cos Ɵ t

y/x= [(u sin Ɵ)t+1/2 gt^2]/(u cos Ɵ t)

simplified it.. and I ended up with

y/x= tan Ɵ + (gt^2)/(2u cos Ɵ t)

So yeah, I'm stuck here.

Help please?

Not sure whether I'm on the right track..

 

[PandaBoy]

2. Yes you are on the right track. You already found your k, its only proving the other part of the equation left. You might need some basic trigonometric identities to finish your proving.

 

[articulate]

How do I continue on? I've tried and I came up with something but I don't think I did it right.

By comparing both equations and using the y=mx+c concept

eq 1: y/x=[g(1+k^2)x]/(2u^2) + k

eq 2: y/x= tan Ɵ + (gt^2)/(2u cos Ɵ t)

So..

tan Ɵ= k

and

[g(1+k^2)]/(2u^2) = (gt^2)/(2u cos Ɵ t)

?

I don't think this is right at all, is it? I shouldn't comparing this:
[g(1+k^2)]/(2u^2) = (gt^2)/(2u cos Ɵ t)

because x is not present in my second equation.

ANYWAY

I just continued on

by comparing this

1+k^2 (from equation 1)

with

t (from equation 2).

Since k= tan Ɵ and t=1+k^2 (not sure this is right at all),

t= 1 + (tan Ɵ)^2


Can someone explain this^ to me? I'm not getting it.

 

[articulate]

Ok. Never mind. Solved!

Just posting this for anyone who's interested..

So..

x= u cos Ɵ t

Thus,

t=x/(u cos Ɵ)... eq1

y= u sin Ɵ t + (gt^2)/2... eq2

Substitute the t in eq2 with eq1, you'll get this:

y= x tan Ɵ + (x^2 g)/(2 u^2 cos^2 Ɵ)


y/x= tan Ɵ + [(x g)(1 + tan^2 Ɵ)]/ (2 u^2)


y/x= x (g(1 +tan ^2 Ɵ))/2u^2 +k

y/x= x[g(1+k^2)]/2u^2 +k

^^(shown)

tan Ɵ=k

 

[rwisz]

Hi there,

The projectile/trajectory equation you have been given is known as the range equation. It is used to calculate the horizontal distance (x) that a projectile will travel given its initial velocity (u), launch angle (Ɵ), and gravitational acceleration (g). The constant k in this equation represents the effects of air resistance on the projectile's motion.

To answer your questions:

1. K is indeed air resistance, also known as drag. It is a force that acts in the opposite direction of an object's motion through a fluid (in this case, air). It is derived from the drag equation, which takes into account the density of the fluid, the object's surface area, and its velocity. In the range equation, k is included to account for the decrease in horizontal distance due to air resistance.

2. The range equation can be derived using principles of projectile motion and basic calculus. It is a combination of the equations for horizontal and vertical displacement, and takes into account the effect of air resistance.

Your approach to simplifying the equation is correct. The final form of the range equation is y/x = tan Ɵ + (k/2) * (x/u)^2. This equation can be rearranged to solve for any of the variables (u, Ɵ, x, or y) depending on what information you have.

I hope this helps clarify the range equation and its use in projectile motion. Good luck with your practical!